Were I to fall towards a typical black hole, the tidal forces would rip me apart well before I got to the event horizon. However, if a black hole were big enough, I could enter the event horizon before tidal forces ripped me apart. How big would this black hole have to be in terms of mass and Schwarzschild radius?
Unless you have better numbers, let's say I don't want to suffer more than 10 Gs of tidal force.
This problem is dealt with (in the context of classical General Relativity) nicely in Taylor & Wheeler's book "Exploring black holes: An introduction to General Relativity" (2000, Addison, Wesley, Longman).
In the section entitled "Project B: Inside the black hole" they perform a calculation for a free-falling observer, based on the Schwarzschild metric for non-rotating black holes, for the time it will take from being "uncomfortable" to reaching the singularity at the centre and the radius at which this occurs.
It turns out that this time is independent of the mass of the black hole and is equal to $$ \tau = \frac{2c}{3}\left(\frac{\Delta r}{g}\right)^{1/2},$$ where $\Delta r$ is your height and $g$ is the differential acceleration you are going to experience between head and feet. The radial coordinate at which this occurs does depend on black hole mass $M$ and is given by $$ r = \left( \frac{2GM \Delta r}{g}\right)^{1/3}$$
So, if we equate the latter with the Schwarzschild radius $r_s = 2GM/c^2$, then the tidal "ripping" (!) takes place prior to reaching the event horizon if the black hole mass is less than $$ M < \frac{c^3}{2G} \left(\frac{\Delta r}{g}\right)^{1/2}$$
This appears to be precisely the result obtained by Alan Rominger using Newtonian gravity! In fact, calculating the radial tidal forces using a Newtonian approximation turns out to be correct for a radially free falling body in the Schwarzschild metric.
If we let $\Delta r=2$ m and $g = 100$ m/s$^2$, then $M<2.86\times 10^{34}$ kg (or $1.43\times10^{4}$ solar masses). More massive than this and (according to classical GR) you would be torn apart after falling inside the event horizon but before reaching the singularity, because the tidal forces grow as $r^{-3}$.
Taking this as a matter of Fermi estimation, I will take the Newtonian form of gravity. No, this isn't great accuracy, but if anyone has any severe theoretical issues to raise, I will be glad to hear them. I will assume that your body extends 1 m out from its center of mass and that the extremities there will experience 10 g before your fingernails bleed and you are pronounced dead.
$$ r_s = \frac{2 GM }{c^2 } \\ \text{ tidal } = 10 g = 2 \frac{ GM}{r_c^3 } \Delta r \\ M = \frac{c^3}{2 G} \sqrt{ \frac{ \Delta r }{ 10 g } } $$
Google can calculate this. I obtain 2e34 kg, or 10,250 solar masses. That wouldn't be the largest black hole in the Milky Way. But still large enough so that finding it would be uncommon compared to the much larger stellar mass category, all of which will kill you while our telescopes can still observe it.
Answer to question Version 1: No one knows. We can answer this question using general relativity to give a classical description but I think there is now serious doubt that GTR describes the inside of a black hole (i.e. within the event horizon) accurately and that we shall need a full quantum theory of gravity to know what happens there.
But the classical description is as follows.
You cannot fall into a black hole without dying: you will hit the singularity: one of the crucial characteristics of the event horizon is that the future of any world line beginning at any point inside the event horizon is a collision with the singularity. But the black hole might be big enough that you have a normal lifespan to live before you get there. Let's look at this further.
See my answer here, where I talk about world lines within a black hole using the really neat Kruskal–Szekeres co-ordinates. Notwithstanding their fearsome name and appearance, their crucial, neat, intuitive property is this: light cones on a KS chart look exactly like they do in flat Minkowsky spacetime.
So, looking at the KS diagram in my other answer, the time you have, depending on your initial speed and other factors, is of the order of a few $G\,M$, where $M$ is the black hole's Schwarzschild mass parameter: equal to half the Schwarzschild radius, thus equal to $G\,M/c^2$, or $G\,M/c^3$ expressed as a time. So, say we want this time to be of the order of $10^9$ seconds: a significant fraction of a human lifetime. This implies a black hole colossal mass of $10^9\,c^3/G$. If I have gotten my conversion from natural to SI units right, this comes out to be $3.9\times 10^{44}{\rm kg}$ or roughly $10^{14}$ solar masses. The Schwarzschild radius will thus be of the order of a human lifetime times a lightyear. By way of comparison, the black hole at our galaxy's center is a paltry four million solar masses. My estimate is, however, significantly below the estimate of the Universe's total energy, so it is in theory possible.
